Question

Consider a launch vehicle of mass 10 tons being launched vertically. The vehicle has 8 tons of propellant, which burns completely at a constant rate over 50 s. If the engine specific impulse is 250 s, and the acceleration due to gravity at sea level is \(g_0\), the acceleration experienced by the vehicle at lift-off is ________.

• A. \(g_0\)
• B. \(2g_0\)
• C. \(3g_0\)
• D. \(4g_0\)

Hint:

First find the thrust from \(F=\dot m I_{sp} g_0\), then subtract the vehicle's own weight from the thrust before dividing by mass.

Answer 1

The Correct Option is C

Step 1: Work everything in terms of the thrust-to-weight ratio instead of raw forces. Define the thrust-to-weight ratio at lift-off as \[ \frac{T}{W_0} = \frac{F}{m_0g_0} \] This dimensionless number directly tells you, in units of $g_0$, how hard the engine is pushing compared to how heavy the vehicle is.

Step 2: Compute the thrust using the specific impulse. Mass flow rate: $\dot m = 8000\text{ kg}/50\text{ s} = 160$ kg/s. Thrust: $F = \dot m I_{sp} g_0 = 160\times250\times g_0 = 40000\,g_0$ N.

Step 3: Compute the thrust-to-weight ratio. \[ \frac{T}{W_0} = \frac{40000\,g_0}{10000\,g_0} = 4 \] This ratio of 4 means the engine can, on its own, support 4 times the vehicle's weight.

Step 4: Convert the thrust-to-weight ratio into a net acceleration. At lift-off the vehicle is not yet moving, so the only two forces are thrust (up) and weight (down). Newton's second law in g units simplifies to \[ \frac{a}{g_0} = \frac{T}{W_0} - 1 \] because the minus 1 is exactly the vehicle's own weight pulling back against the thrust ratio. \[ \frac{a}{g_0} = 4 - 1 = 3 \implies a = 3g_0 \] This thrust-to-weight approach is the same idea rocket engineers use to size a launch vehicle's engines, and it gives the identical answer as the direct force-balance calculation. \[ \boxed{a = 3g_0} \]